Question

Two point bodies of masses $m$ and $3m$ are connected by a massless spring of spring constant $k = m\omega_0^2$ and kept on a frictionless horizontal surface. The spring is extended by a small distance $l$ over its natural length at time $t = 0$ and then released so that the masses execute simple harmonic motion. The maximum speed of the particle with mass $m$ is given by

• A. $\frac{\sqrt{3}\omega_0 l}{2}$
• B. $\frac{2\omega_0 l}{\sqrt{3}}$
• C. $\frac{\omega_0 l}{\sqrt{3}}$
• D. $\frac{3\omega_0 l}{4}$

Hint:

Using conservation of momentum and conservation of energy at the point of maximum speed (equilibrium position) yields the same result:
\[ \frac{1}{2} k l^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (3m) v_2^2 \]
With $m v_1 = 3m v_2 \implies v_2 = v_1 / 3$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a two-body oscillator system. Both masses execute simple harmonic motion about their common center of mass (COM) on a frictionless surface.

Step 2: Key Formula or Approach:
- The reduced mass $\mu$ of a two-body system is:
\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]
- The angular frequency of oscillation is:
\[ \omega = \sqrt{\frac{k}{\mu}} \]
- In the COM frame, the displacement of mass $m_1$ is related to the extension of the spring by:
\[ x_1 = \frac{m_2}{m_1 + m_2} x \]

Step 3: Detailed Explanation:

• First, let us find the reduced mass of the system:
\[ \mu = \frac{m \cdot 3m}{m + 3m} = \frac{3m^2}{4m} = \frac{3}{4}m \]

• The angular frequency of oscillation is:
\[ \omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{m\omega_0^2}{\frac{3}{4}m}} = \frac{2}{\sqrt{3}}\omega_0 \]

• Since the center of mass remains at rest, the extension $x$ of the spring is shared between the two masses.
The displacement $x_1$ of the mass $m$ is:
\[ x_1 = \frac{3m}{m + 3m} x = \frac{3}{4}x \]

• The maximum extension of the spring is $l$. Therefore, the amplitude of oscillation for mass $m$ is:
\[ A_1 = \frac{3}{4}l \]

• The maximum speed of the mass $m$ is given by:
\[ v_{1,\max} = \omega A_1 \]
\[ v_{1,\max} = \left(\frac{2}{\sqrt{3}}\omega_0\right) \left(\frac{3}{4}l\right) \]
\[ v_{1,\max} = \frac{6}{4\sqrt{3}}\omega_0 l = \frac{3}{2\sqrt{3}}\omega_0 l = \frac{\sqrt{3}}{2}\omega_0 l \]



Step 4: Final Answer:
The maximum speed of the particle with mass $m$ is $\frac{\sqrt{3}\omega_0 l}{2}$.